VITEEE 2010 Paper

y=sin−1(

x2

1+x2

)
For y to be defined |

x2

1+x2

|<1
which is true for all x ∊ R.
Now, y=sin−1(

x2

1+x2

)
⇒

x2

1+x2

=sin‌y
⇒ x=

√sin‌y

1−sin‌y

For the existence of x
sin y ≥ 0 and 1 - sin y > 0
⇒ 0≤sin‌y<1⇒0≤y<

π

2

Thus, range of the given function is [0,

π

2

).